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          【LeetCode】面试题 08.11之硬币组合问题
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        <h4 id="题目描述："><a href="#题目描述：" class="headerlink" title="题目描述："></a>题目描述：</h4><p>给定数量不限的硬币，币值为25分、10分、5分和1分，编写代码计算n分有几种表示法。<a id="more"></a>(结果可能会很大，你需要将结果模上1000000007)</p>
<h4 id="示例："><a href="#示例：" class="headerlink" title="示例："></a>示例：</h4><pre>
输入: n = 5
 输出：2
 解释: 有两种方式可以凑成总金额:
5=5
5=1+1+1+1+1
</pre>

<pre>
输入: n = 10
 输出：4
 解释: 有四种方式可以凑成总金额:
10=10
10=5+5
10=5+1+1+1+1+1
10=1+1+1+1+1+1+1+1+1+1
</pre>

<h4 id="分析："><a href="#分析：" class="headerlink" title="分析："></a>分析：</h4><p>这个题一开始我是没一点头绪的，简直是无从下手，不得不参考一下官方解答。不过最后看了官方一大篇的推导，我似乎更加蒙了。最后还是反反复复地看代码，才慢慢明白一点了。</p>
<p>其实用给定这四种面值的硬币组合不同的面值，换个说法其实就类似于一个列举所有可能的过程，我们可以对不同面值的硬币进行单独的运算，最后把各种面值的组合情况进行累加就可以；其中最主要的难点是如何解决金额不是硬币面值的整数倍问题。</p>
<p>例如：我们要计算金额为11，使用面值为5的硬币如何组成；很显然仅仅使用面值为5的硬币是不可能组成金额11的，我们可以换个角度思考，计算用面值5的硬币组成金额11，其实就相当于必须用<code>11 / 5 = 2</code> 个面值为5的硬币，而余数<code>11 % 5 = 1</code>是小于5的，不能继续使用面值5的硬币，所以只能由小于面值为5的硬币组成。</p>
<p>从这个例子其实就不难看出，想要计算面值大一点的硬币组成总金额，必须要先计算出面值小的硬币的组合方式。因此，在使用循环列举的时候，我们要先计算用面值最小的硬币组成总金额。</p>
<p>另外，由于要组成的金额是未知的，也就是说用不同面值的硬币去组合的时候，遇到的余数肯定是各种各样的，因此我们必须计算出从金额<code>1</code>到<code>n</code>的所有金额的组合情况。</p>
<h4 id="Java实现"><a href="#Java实现" class="headerlink" title="Java实现"></a>Java实现</h4><figure class="highlight angelscript"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line">&lt;pre&gt;</span><br><span class="line"><span class="keyword">class</span> <span class="symbol">Solution</span> &#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="built_in">int</span> waysToChange(<span class="built_in">int</span> n) &#123;</span><br><span class="line">        <span class="built_in">int</span>[] arr = new <span class="built_in">int</span>[n + <span class="number">1</span>]; <span class="comment">//arr[i]表示面值为i时的组合方式</span></span><br><span class="line">        arr[<span class="number">0</span>] = <span class="number">1</span>; <span class="comment">//方便计算</span></span><br><span class="line">        <span class="built_in">int</span>[] coins = &#123;<span class="number">1</span>, <span class="number">5</span>, <span class="number">10</span>, <span class="number">25</span>&#125;;</span><br><span class="line">        <span class="keyword">for</span>(<span class="built_in">int</span> i = <span class="number">0</span>; i &lt; <span class="number">4</span>; i++)&#123;</span><br><span class="line">            <span class="built_in">int</span> coin = coins[i];</span><br><span class="line">            <span class="keyword">for</span>(<span class="built_in">int</span> j = coin; j &lt;= n; j++)&#123;</span><br><span class="line">                arr[j] = (arr[j] + arr[j - coin]) % <span class="number">1000000007</span>;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> arr[n];</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line">&lt;/pre&gt;</span><br></pre></td></tr></table></figure>


<p>虽然代码数量看起来不多，但是这里面的<code>arr</code>数组的设计是非常巧妙的，非常值得学习借鉴。</p>

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